Question

Set up a double integral to find the volume of the solid bounded by the graphs of the equations. \(x^{2}+z^{2}=1, y^{2}+z^{2}=1,\) first octant

Short answer

The double integral that gives the volume of the solid bounded by the graphs is \(\int_{0}^{\pi/2} \int_{0}^{1} r \, dr \, d\theta\).

Step-by-step solution

Conversion to Polar Coordinates

Convert the given equations from cartesian to cylindrical coordinates. We can use polar coordinates \((r,\theta)\) because we know that \(x = r\cos(\theta)\), \(y = r\sin(\theta)\), and \(z = r\). The equations then become \(r^{2}\cos^{2}\theta + r^{2} = 1\) and \(r^{2}\sin^{2}\theta+ r^{2} = 1\).

Solving the Equations

Resolve the equation to find the limits for the double integral. Solve both equations to get \(r^{2}=1\). This indicates that \(r\) ranges from 0 to 1.

Determine the range for \(\theta\)

It is stated that we are looking at the first octant, which means \(\theta\) ranges from 0 to \(\pi/2\).

Set up the Double Integral

Combine all findings to set up the double integral. To calculate the volume, we integrate over the region of interest. This results in the double integral: \(\int_{0}^{\pi/2} \int_{0}^{1} r \, dr \, d\theta\). The polar elements are multiplied by \(r\) because in polar coordinates, a small area is given by \(r \, dr \, d\theta\). The integral \(r \, dr\) calculates the area of each of those small area elements, and integrating over \(r\) and \(\theta\) sums up all of those small areas to give the total volume of the solid.

Key concepts

Cylindrical Coordinates

To understand volume calculation in math, one needs to grasp the concept of coordinate systems. Cylindrical coordinates offer a three-dimensional extension of polar coordinates. They are particularly useful for solving problems that involve symmetry around a central axis. In this system, a point in space is described by three values: the radial distance from the origin \( r \), the angle \( \theta \) from the positive x-axis, and the height \( z \) above the xy-plane.

In our exercise, the cylindrical coordinates simplify the equations of the two surfaces that bound the solid. With the transformations \( x = r\cos(\theta) \), \( y = r\sin(\theta) \) and \( z \), we define every point in the volume using this coordinate system. This facilitates the computation of the integral, especially when the volume has rotational symmetry, which is often the case for problems limited to the first octant.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system in which each point on a plane is defined by a distance from a reference point and an angle from a reference direction. This system is particularly suitable for dealing with problems involving circular or rotational symmetry.

In the context of the given problem, the cylindrical coordinates condensed the double integral computation into a polar coordinates format, as the problem deals with circular cross-sections parallel to the xy-plane. By recognizing that \( r^{2} = x^{2} + y^{2} \), we were able to rewrite our boundaries in terms of \( r \) and \( \theta \) rather than \( x \) and \( y \). This greatly simplifies the calculation of the volume, especially when integrating over circular regions or sectors as in the first octant.

Volume of Solid

Calculating the volume of a solid can be approached in various ways depending on the shape and symmetry of the solid in question. A double integral is a powerful tool for finding the volume of more complex shapes — particularly those that aren't easily decomposable into simple geometric objects.

The double integral works by accumulating 'slices' of volume. In polar or cylindrical coordinates, the integral includes an extra \( r \) term that accounts for the 'stretching' effect of converting a flat area element into the curved space of polar coordinates. For the solid described in the problem, the double integral calculates the total volume by integrating the area of infinitely many small circular rings over the specified region.

First Octant

The first octant in three-dimensional space is the portion where all the coordinates \( x \), \( y \), and \( z \) are positive. It's akin to the first quadrant of the xy-plane but extended into 3D space.

Identifying the region of integration is crucial when setting up the bounds for calculating volumes using integrals. Our problem specifies the solid's location within the first octant, which simplifies setting up the limits for \( r \) and \( \theta \) since \( \theta \) only varies from \( 0 \) to \( \pi/2 \) and \( r \) is only positive (and bounded, in our case, by the unit circle). While this restriction might seem simple, it substantially aids in defining the limits of our double integral and guides us to the correct solution.